IF 300√3m HIGH TOWER MAKES ANGLE OF ELEVATION AT A POINT ON GROUND WHICH IS 300M AWAY FROM ITS FOOT THEN FIND THE ANGLE OF ELEVATION.
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CD is the height of the flag and suppose the height of tower BC = h
Now consider right triangle ABD , we have;
tan 60° = BDAB = BD120⇒3√ = BD120⇒BD = 1203√
Also considering right triangle ABC, we have;
tan 45° = BCAB = h120⇒1 = h120⇒h = 120
So height of flag = BD - BC = 1203√−120 = 120(3√−1) = 120×0.732 = 87.84
Therefore height of the flag is 87.84 m.
The values are changed but the solution is same.
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Now consider right triangle ABD , we have;
tan 60° = BDAB = BD120⇒3√ = BD120⇒BD = 1203√
Also considering right triangle ABC, we have;
tan 45° = BCAB = h120⇒1 = h120⇒h = 120
So height of flag = BD - BC = 1203√−120 = 120(3√−1) = 120×0.732 = 87.84
Therefore height of the flag is 87.84 m.
The values are changed but the solution is same.
Hope the answer was helpful to you If it was please mark it as brainlist ❤️
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3
60°
Step-by-step explanation:
The angle of elevation is 60°
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